Problem
Primes Complexity
18-1
Complexity
18-2
The Instance: A positive integer k.
Primes
Question: Is k prime?
The complement of Primes, the Composite problem, belongs to NP. Therefore Primes is in coNP Recently M.Agarwal et al. Proved that Primes can be solved in polynomial time (see http://www.cse.iitk.ac.in/news/primality.html) Complexity Andrei Bulatov However, the probabilistic algorithm we are going describe is far more efficient
Complexity
18-3
Complexity
18-4
Residues
For a positive integer n, we denote
Z n the set {0,1,2,,n 1}
Z + the set {1,2,,n 1} n
Complexity of Arithmetic
Given two integers, a and b, we can compute a + b in O(max(log a, log b)) a b in O(log a log b)
a b cannot be computed in polynomial time, because the size of this number is blog a
It is possible modulo n
+,, x y addition, multiplication and exponentiation modulo n
Z n together with these operations is called the set of residues modulo n
Let b0 b1b2 K bk be the binary representation of b (k = log b) Then b = b0 2 0 + b1 21 + K + bk 2 k that implies
Every integer m, positive or negative, has a corresponding residue m mod n For example, 17 mod 5 = 2 20 mod 5 = 0 -1 mod 5 = 4
a b (mod n ) = a b0 2 a b1 2 K a bk 2
0
1
k
First, we consecutively compute a 2 , a 2 ,K, a 2 in O ( k log2 n ) Then we compute the product again in O ( k log2 n )
0
1
k
Complexity
18-5
Complexity
18-6
Prime and Coprime Fermats Theorem
Integers a and b are called coprime if their greatest common divisor is 1 Theorem (Fermats Little Theorem) For example, 16 and 27 are coprime, and 15 and 18 are not Theorem (Chinese Remainder Theorem) If p and q are coprime then, for any a and b, there is x such that x a (mod p )
x b(mod q )
If p is prime then, for any a Z + we have a p1 1(mod p ) p
If the converse were true, we could use it for a probabilistic primality test: Choose k residues modulo n; Compute their n 1 powers;
For example, if p = 5, q = 3, and a = 2, b = 1, then x can be chosen to be 7
Accept if all results are 1 (mod n), reject otherwise
1
Complexity
18-7
Complexity
18-8
Carmichael Numbers
Unfortunately, the converse is true just almost Definition A number n passes Fermats test if a n1 1(mod n) for all a coprime with n A number that passes Fermats test is called pseudo-prime
Roots of 1
A square root of 1 modulo n is a number a such that a 2 1(mod n ) Clearly, 1 and -1 (that is n 1) are always roots of 1, but if n is composite, then it may have more than two roots of 1 For example,
8 has four roots of 1: 1, -1, 3, and 5
+ One can straightforwardly check that, for any a 561 Z , coprime with 561, a 560 1(mod 561)
561 has eight: 1, -1, 188, 373 (find the remaining four)
561 is a Carmichael number
n is said to be a Carmichael number if, for any prime divisor p of n, p 1 | n 1
Pseudo-prime = Prime + Carmichael
Lemma Any Carmichael number has at least 8 roots of 1
Complexity
18-9
Complexity
18-10
Algorithm
On input n if n is even, then if n = 2 accept, otherwise reject
+ select randomly a1 , a 2 ,K, a k Z n
Analysis
First we show that the algorithm does not give false negatives, that is it accepts all prime numbers If n = 2 then n is accepted. Let n be an odd prime number
for i = 1 to k do - if a
n 1 i
1(mod n ) then reject
- let n 1 = st where s is odd and t = 2h is a power of 2 - compute the sequence ais2 , ais2 ,K, ais2
0 1 h
Then n passes Fermat test modulo n
- if ais2 1 then
j
n cannot be rejected in the last line, because n has only two roots of 1
let j be the maximal with this property if ais2 1 then reject
j
accept
Complexity
18-11
Complexity
18-12
Next we show that if n is composite, then Pr[n accepted] 2 k A number a Z + such that a does not pass either Fermat test or the n square root test, is called a witness It is enough to prove that Pr[a is a witness] 1/2, or, in other words, that at least half of the elements of Z + are witnesses n For every nonwitness d we find a witness d such that if d1 d 2 then d '1 d ' 2 For a nonwitness a the sequence a s2 , a s2 ,K, a s2 1s only, or it contains -1 followed by 1s
0 1 h
Let d be a nonwitness of the second type such that the 1 app...